# What does the “closed over”/“closed under” terminology mean exactly and where did it come from?

I've been trying to teach my partner some set theory, and I got thrown for a loop while trying to give her a precise definition of some basic terminology. So we've heard of a set being described as "closed under" an operation, as well as an operation being "closed over" a set.

First, correct me if I'm wrong, but my impression is that these things mean the same thing.

Second, does this terminology have a precise definition? For instance we'd normally say that set $A$ is closed under $f:X\rightarrow Y$ if $A\subseteq X$ and for any $x\in A$, $f(x)\in A$. But it's also common to extend the definition such that $X$ isn't a superset of $A$ but of $A^n$ for some $n$. For instance, we'd say $A$ is closed under addition if the addition operator maps elements of $A^2$ to $A$. Does this extend to anything bigger than $n$-tuples? Infinite sequences, for instance?

Third, what's the motivation for this terminology and is it in any way related to the normal topological definition of closedness? There's a way that closed sets are related to sets that are closed under a particular operation in certain topological spaces (if you allow the usage that includes infinite sequences as above), but I haven't come up with a general relation between the two concepts. What's the motivation for saying an operation is closed over a set though, or is that just a corruption of the former? Does anyone know the historical reasoning behind any of this terminology?

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I have never heard anyone use "closed over." – Qiaochu Yuan Jul 14 '11 at 19:41
I've seen it used, but not nearly as much as the alternative. Googling suggests that mostly programmers use that terminology (and that it's common and means something different in functional programming.) So presumably it's just a rare corruption of the original terminology. – jsn Jul 14 '11 at 19:49
I'm wondering about the need for [set-theory] and [general-topology]. – Asaf Karagila Jul 14 '11 at 20:00
Sorry if I've cast an unnecessarily wide net: this my first question, I don't know what the norms are for tags. I figured the question came up in the context of giving someone an introduction to set theory and general topology, so people who knew about/were interested in those things would be most likely to be able to answer my question. – jsn Jul 14 '11 at 20:05
Since the answers given so far are more general than set theory or topology topics, I will remove the tags. I am not 100% that [logic] is needed and will be glad to have the decision to leave it out at the moment reconsidered by other people as well. – Asaf Karagila Jul 14 '11 at 20:38

In an ordinary mathematical context I would never say that an operation $\mathcal O$ is closed over a set $A$; I consider that an error for ‘$A$ is closed under $\mathcal O$’. The latter terminology can be properly applied very generally. For starters, if $A \subseteq S$, $n \in \omega$, and $f:S^n \to S$, ‘$A$ is closed under $f$’ means precisely that for every $\langle a_1,\dots,a_n \rangle \in A^n$, $f(a_1,\dots,a_n) \in A$.

This is the most common algebraic usage, I think, but it’s just the beginning. For instance, the non-negative integers $n$ can be replaced by any ordinal $\alpha$, and the operation need not be defined on all members of $S^\alpha$. Suppose that $X$ is a topological space and $A \subseteq X$. ‘$A$ is closed under (the operation of taking) limits of convergent sequences’ then means that if $\langle a_n:n \in \omega \rangle$ is a sequence of points in $A$ that converges as a sequence in the space $X$, its limit point is actually in $A$. Here $\alpha = \omega$, and the operation is defined only for those elements of $X^\omega$ that converge in $X$. This example at least starts to show the relationship between the general notion and that of topological closedness.

Moreover, the terminology is still used when the input to the operation isn’t ordered: it’s perfectly correct to say that a family $\mathcal A$ of sets is closed under (taking) finite intersections, for instance, meaning that if $\mathcal F$ is any finite subfamily of $\mathcal A$, $\bigcap \mathcal F \in \mathcal A$. If $S$ is the underlying set, the operation is $\mathcal O:[(\mathcal P(S)]^{< \omega} \to \mathcal P(S):\mathcal F \mapsto \bigcap \mathcal F$, and $\mathcal A \subseteq \mathcal P(S)$ is closed under it because $\mathcal O$ maps $[\mathcal A]^{< \omega}$ into $\mathcal A$. (Here $[X]^{< \omega}$ denotes the set of finite subsets of $X$.)

It would be a bit difficult to formulate an exhaustive formal definition of the usage, and I’m not at all sure that it would be particularly helpful; it seems more useful to present a variety of examples showing the flexibility of the usage.

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Ah, thanks. I had the operation that takes convergent sequences to their limits in mind when I wrote the third part of my question, but I was weirded out by it not being defined for all sequences. And my thought was that if we could extend this to nets or filters or whatever, there'd be a general correspondence. I hadn't considered the usage "closed under finite intersections", which makes me agree that the usage is probably more broad and less sharply defined than I figured. – jsn Jul 15 '11 at 20:38
+1 for the last sentence. – Nate Eldredge Jul 15 '11 at 21:34

I don't know the history of the terminology, but "closed under" does have a relation to the notion of "closed" in topology. The "closed over" terminology is probably a hypercorrection due to wrongly interpreting "under" as a preposition denoting position.

We say that some collection $S$ is closed under some $k$-input operation $f$ if the result of applying $f$ to any inputs from $S$ remains in $S$. If $S$ is not closed under $f$, we could form the closure of $S$ under $f$, defined as $\bigcup_{n=0}^\infty S_n$ where $S_0 = S$ and $S_{n+1} = S_n \cup Im_f({S_n}^k)$. It is easy to see that the closure of $S$ under $f$ will be closed under $f$.

In general, we can talk about the closure of a collection $S$ under some more general operation $F$ that takes as input a collection and returns as output another collection, where the closure $cl_F(S)$ is defined as $\bigcup_{n=0}^\infty S_n$ where $S_0 = S$ and $S_{n+1} = S_n \cup F(S_n)$. The above more specific definition is then just a special case where $F(X) = \{ f(\vec{x}) : \vec{x} \in X^k \}$. It turns out that $cl_F(S)$ is the smallest collection that contains $S$ and is closed under $F$.

This general notion of closure is used everywhere in mathematics, such as:

• The linear span of a set of vectors (closure under addition and scalar multiplication)

• The algebraic closure of a field (closure under adding roots of polynomials, more or less)

• The transitive closure of a partial order (closure under adding $x \prec z$ for every $x \prec y \prec z$)

• The σ-algebra of a collection of sets (closure under complement and countable union)

• Topological closure of a set (which I'm going to say more about below)

The topological closure in a metric space is the closure under taking limit points. As mentioned, it is also the smallest superset that is closed under taking limit points. In an arbitrary topological space we have no metric, but we can use the topological definition of limit points, and the same holds. It turns out that in this case just one step of adding limit points already produces the closure.

So one can say that the topological closure is a specific instance of the general notion of closure.

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First, closed under and closed over mean the same thing.

Second, your definition is correct except its usually done with an operation. An operation $*$ is a function from $G\times G \to H.$ An operation is closed over a set $G$ if if an only if, for every $x,y\in G$ $f(x,y)=x*y\in G.$

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My impression was that: the operation could be defined on a superset of $G\times G$ and we'd still stay that $G$ is closed under the operation if the image of $G\times G$ was a subset of $G$. Is this not correct? Second, am I to take it that the primary usage here is for binary operations (not operations of arbitary arity, as I assumed above)? – jsn Jul 14 '11 at 19:52